oracle面试题求解

2012-02-03 02:15:23 404 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

AI 赋能测试全生命周期：多 Agent 协同的自动化提效方案本文提出了一种基于Coze平台的多Agent协同测试解决方案，旨在解决软件测试中PRD解析不清晰、测试点覆盖不全、用例编写效率低、代码生成耗时等痛点。通过PRD优化、测试点生成、测试用例生成和单元测试框架生成四个Agent的协同工作，结合Cursor工具实现从需求到代码的全链路自动化。以电商平台登录模块为例，展示了从模糊PRD到标准化测试用例和可执行Java单元测试代码的完整转化流程。该方案通过AI赋能显著提升测试效率，减少人工干预，实现测试覆盖率≥95%，为快速迭代的软件研发提供了高效的测试自动化支持。阅读详情

GOODS(货品表)

GOODSID(主键) GOODSNAME MEMO

1 青霉素

2 西瓜霜

3 创可贴

4 西洋参

SU(进货表)

GOODSID(主键) SUQTY

1 60

2 70

SA(销售表)

GOODSID(主键) SAQTY

1 80

2 90

问题一：进货记录，给出SQL达到如下效果

GOODSID GOODSNAME SUQTY

1 青霉素 60

2 西瓜霜 70

3 创可贴 0

4 西洋参 0

问题二：进销对比，给出SQL达到如下效果

GOODSID GOODSNAME SUQTY SAQTY

1 青霉素 60 0

2 西瓜霜 70 80

3 创可贴 0 90

(有会的请留下答案)

------------------------------------------------

sql优化问题

select * from sa where sa.G

...全文

华为Atlas 300I推理卡驱动安装全攻略：从环境检查到常见问题解决本文提供了华为Atlas 300I推理卡驱动安装的完整实战指南。从安装前的系统环境检查、内核版本匹配，到详细的二进制.run文件安装步骤、源码编译及高级定制化部署，涵盖了驱动安装的全流程。文章重点解析了安装过程中的关键决策、常见问题排查与解决方案，并针对生产环境的多卡配置、性能调优及安全加固给出了专业建议，旨在帮助用户高效、稳定地完成AI推理加速卡的部署。阅读详情

麦田

领域专家: 后端开发技术领域

Oracle 17,133 成员 55,258 内容

+ 关注 + 加入社区

切换为时间正序

buryMyLove 2012-02-10

打赏
举报

第二题LZ搞错了，根据第二题的答案，销售表的两个id应该是2，3而不是1，2

第一题：
select a.*,nvl(b.SUQTY) from GOODS a,SU b where a.GOODSID=b.GOODSID(+);

第二题是因为西洋参没有进也没有销所以结果里不出么，还是说进销相抵以后库存是0？不出的原因不同sql的写法也不同，如果是因为西洋参的进销都没有所以不出这条数据：
select a.*,nvl(b.SUQTY)，nvl(c.SAQTY) from GOODS a,SU b,SA c
where a.GOODSID=b.GOODSID(+)
and a.GOODSID=c.GOODSID(+)
and a.GOODSID in (select GOODSID from SU union select GOODSID from SA);

jmt1111 2012-02-09

打赏
举报



with goods as (

select 1 goodsid,'青霉素' goodsname from dual

union all

select 2 goodsid,'西瓜霜' goodsname from dual

union all

select 3 goodsid,'创可贴' goodsname from dual

union all

select 4 goodsid,'西洋参' goodsname from dual

)

, su as (

select 1 goodsid,60 suqty from dual

union all

select 2 goodsid,70 suqty from dual

),sa as (

select 2 goodsid,80 saqty from dual

union all

select 3 goodsid,90 saqty from dual

)



--select a.goodsid,goodsname,nvl(suqty,0) from goods a full outer join su b on (a.goodsid=b.goodsid);



--select a.goodsid,goodsname,nvl(suqty,0),nvl(saqty,0) 

--from goods a full outer join su b on (a.goodsid=b.goodsid) full outer join sa c on (a.goodsid=c.goodsid)

--where a.goodsid=b.goodsid or a.goodsid=c.goodsid;

mingchaoyan 2012-02-09

打赏
举报

[Quote=引用 4 楼 phoenix_99 的回复:]

SQL code

with goods as (
select 1 goodsid,'青霉素' goodsname from dual
union all
select 2 goodsid,'西瓜霜' goodsname from dual
union all
select 3 goodsid,'创可贴' goodsname from dual
union all
select 4 good……
[/Quote]

漂亮呀！！！

merrill 2012-02-09

打赏
举报

问题二：进销对比，给出SQL达到如下效果

GOODSID GOODSNAME SUQTY SAQTY

1 青霉素 60 0

2 西瓜霜 70 80

3 创可贴 0 90
问题二应该是楼主搞错了

问题一：
select a.goodsid, a.goodsname, nvl(b.suqty, 0)
from GOODS a, SU b
where a.goodsid = b.goodsid(+);

陈大鱼 2012-02-09

打赏
举报

[Quote=引用 6 楼 andhai 的回复:]
create table su
(goodsid char(1),suqty number(3,0));

insert into su values('1',60);
insert into su values('2',70);

create table sa (goodsid char(1), saqty number(3,0));

insert into sa value……
[/Quote]

第二个应该改成

select goods.goodsname goodsid ,goods.memo,nvl(su.suqty,0),nvl(sa.saqty,0)
from goods,su,sa
where goods.goodsname= su.goodsid(+)
and goods.goodsname-1= sa.goodsid(+)
and rownum<4
order by goodsname;

huxingz 2012-02-09

打赏
举报

答案1:
SELECT G.GOODSID, G.GOODSNAME, NVL(SUM(S.SUQTY), 0) AS SUQTY
FROM GOODS G, SU S
WHERE G.GOODSID = S.GOODSID(+)
GROUP BY G.GOODSID, G.GOODSNAME
答案2:

SELECT G.GOODSID,
G.GOODSNAME,
NVL(SUM(S.SUQTY), 0) AS SUQTY,
NVL(SUM(A.SAQTY), 0) AS SAQTY
FROM GOODS G, SU S, SA A
WHERE G.GOODSID = S.GOODSID(+)
AND G.GOODSID = A.GOODSID(+)
GROUP BY G.GOODSID, G.GOODSNAME
HAVING NVL(SUM(S.SUQTY), 0) + NVL(SUM(A.SAQTY), 0) > 0

另外,如果题目没有出错的话,那么估计是LZ自己写错了,你的结果根本就是看的莫名其妙,请问GOODSID作为主键体现在哪里

andhai 2012-02-07

打赏
举报

create table su
(goodsid char(1),suqty number(3,0));

insert into su values('1',60);
insert into su values('2',70);

create table sa (goodsid char(1), saqty number(3,0));

insert into sa values('1',80);
insert into sa values('2',90);
create table goods
(goodsname char(1),memo varchar2(10));
insert into goods values('1','青霉素');
insert into goods values('2','西瓜霜');
insert into goods values('3','创可贴');
insert into goods values('4','西洋参');

1.
select goods.goodsname goodsid,goods.memo,nvl(su.suqty,0)
from goods,su
where
goods.goodsname= su.goodsid(+)
order by goods.goodsname

2.
select goods.goodsname goodsid ,goods.memo,nvl(su.suqty,0),nvl(sa.saqty,0)
from goods,su,sa
where goods.goodsname= su.goodsid(+)
and goods.goodsname= sa.goodsid(+)

cutebear2008 2012-02-07

打赏
举报

标准的left join！

Phoenix_99 2012-02-03

打赏
举报



with goods as (

select 1 goodsid,'青霉素' goodsname from dual

union all

select 2 goodsid,'西瓜霜' goodsname from dual

union all

select 3 goodsid,'创可贴' goodsname from dual

union all

select 4 goodsid,'西洋参' goodsname from dual

)

, su as (

select 1 goodsid,60 suqty from dual

union all

select 2 goodsid,70 suqty from dual

),sa as (

select 1 goodsid,80 saqty from dual

union all

select 2 goodsid,90 saqty from dual

)



select goods.goodsid,goods.goodsname,nvl(suqty,0) suqty,nvl(saqty,0) saqty from goods,su,sa where 

goods.goodsid=su.goodsid(+) and su.goodsid=sa.goodsid(+)

Phoenix_99 2012-02-03

打赏
举报



with goods as (

select 1 goodsid,'青霉素' goodsname from dual

union all

select 2 goodsid,'西瓜霜' goodsname from dual

union all

select 3 goodsid,'创可贴' goodsname from dual

union all

select 4 goodsid,'西洋参' goodsname from dual

)

, su as (

select 1 goodsid,60 suqty from dual

union all

select 2 goodsid,70 suqty from dual

)

select goods.goodsid,goods.goodsname,nvl(su.suqty,0) suqty from goods,su where goods.goodsid = su.goodsid(+)

order by 1



with goods as (

select 1 goodsid,'青霉素' goodsname from dual

union all

select 2 goodsid,'西瓜霜' goodsname from dual

union all

select 3 goodsid,'创可贴' goodsname from dual

union all

select 4 goodsid,'西洋参' goodsname from dual

)

, su as (

select 1 goodsid,60 suqty from dual

union all

select 2 goodsid,70 suqty from dual

),sa as (

select 1 goodsid,80 saqty from dual

union all

select 2 goodsid,90 saqty from dual

)



select goods.goodsid,goods.goodsname,nvl(suqty,0) suqty,nvl(saqty,0) saqty from goods,su,sa where 

goods.goodsid=su.goodsid(+) and su.goodsid=sa.goodsid(+)

麦田 2012-02-03

打赏
举报

csdn博客原试题http://blog.csdn.net/itmyhome/article/details/7231238

麦田 2012-02-03

打赏
举报

[Quote=引用 1 楼 dzxccsu 的回复:]

真是面试题目？是作业吧
[/Quote]
今天上午刚面试完的，格式有点不好使，看着有点不好看。

dzxccsu 2012-02-03

打赏
举报

真是面试题目？是作业吧

oracle面试题求解

相关推荐